Question

6. + 1/5 points Previous Answers LarLinAlg8 2.5.050. The figure below illustrates an example of a Markov chain with reflecting boundaries. 100% 60% 70% si (S2) (S ) 30% 100% ( SA 40% (a) Explain why it is appropriate to say that this type of Markov chain has reflecting boundaries. When the chain reaches S1 or S4, it is certain in the next step to transition to an adjacent state, S2 or Sz respectively. When the chain reaches S2 or S3, it is certain in the next step to stay at that state. When the chain reaches S1 or S4, it is certain in the next step to stay at that state. When the chain reaches S2 or S3, it is certain in the next step to transition to an adjacent state, S1 or S4 respectively. (b) Use the figure to write the matrix of transition probabilities P for the Markov chain. 0 0.6 0 1 0.3 0 0 .7 P= 0010 (c) Find p30 and p31. Find several other high even powers 2n and odd powers 2n + 1 of P. What do you observe? High even or odd powers of P give similar results where the columns are inverses. High even or odd powers of P give similar results where the rows alternate. • High even or odd powers of P give similar results where the columns alternate. High even or odd powers of P give similar results where the rows are all the same. High even or odd powers of P give similar results where the columns are all the same.

Answer 1

solution 6.) foo 02 P = 0.4 0 0.6 0 10 0.3 00: Pero 0.4 O o ono! OO 0.6 0110 00.6 01 0 0. 7 0 0.30 0.7 . 0:3 p0 .4 0.6 0 0 . 0012 0 0 0 0.+ 03 Co) P p30. 0464 0 0.833 0 7 10 0.4167 0 05833 0-16+ 0 0.8330 lo 6.41676 0-5833) 3.ro 004167 0 0.58337 M167 0 .833 0 0 0.4167 0.5833 0167 0 00833 0 for Ever and odd powers p gives result where the column alternate. d.) for Steady A state matrix P - P . po! 007 10.4 00560 0 0 3 0 0 4 52:51, 0.45,0653752;.. 6.352 + 0 754 25+3, 53 - 54, we have 51 +52 +5₃ +54 21 Solving these we get = 1 1/4 The stead state probabilites au Same It is due the fact that the chain it is fail to converge for a. 1/47 Yu is periodic